Posted by **isa** on Tuesday, February 17, 2009 at 9:41pm.

suppose an arithmetic sequence and a geometric sequence with common ration r have the same first two terms. show that the third term of the geometric series is r^2/(2r-1) tomes the third term of the arithmetic sequence

- math -
**Reiny**, Tuesday, February 17, 2009 at 10:10pm
the first 3 terms of the AS are a, a+d, a+2d

the first 3 terms of the GS are a, ar, ar^2

you said the second terms are equal

then a+d = ar

d = ar-a

so third term of GS/third term of AS

= ar^2/(a+2d)

= ar^2(a + 2(ar-a)

= ar^2/(a + 2ar - 2a)

= ar^2/(2ar - a) now divide top and bottom by a

= r^2/(2r-1)

- math -
**Anonymous**, Sunday, March 1, 2015 at 9:36am
swaggot swag

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